A Level Further Mathematics 9231 Paper 32, Oct/Nov 2024

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Question paper

9231/32 · Oct/Nov 2024

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1

A particle of mass 2 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 100 N. The other end of the string is attached to a fixed point O on a smooth horizontal surface. The particle is moving in a horizontal circle about O with the string taut and with constant angular speed 5 radians per second. Find the extension of the string.

[3]
2

A particle P of mass m is attached to one end of a light elastic spring of natural length a and modulus of elasticity 5mg. The other end of the spring is attached to a fixed point O. The spring hangs vertically with P below O. The particle P is pulled down vertically and released from rest when the length of the spring is 32a. Find the distance of P below O when P first comes to instantaneous rest.

[4]
3

The diagram shows two identical smooth uniform spheres A and B of equal radii and each of mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds 2u and 3u respectively. Immediately before the collision, A’s direction of motion makes an angle θ with the line of centres and B’s direction of motion is perpendicular to that of A. After the collision, B moves perpendicular to the line of centres. The coefficient of restitution between the spheres is 13.

Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a

Find the value of tan⁡θ.

[3]
b

Find the total loss of kinetic energy as a result of the collision.

[2]
c

Find, in degrees, the angle through which the direction of motion of A is deflected as a result of the collision.

[2]
4

The end A of a uniform rod AB of length 6a and weight W is in contact with a rough vertical wall. One end of a light inextensible string of length 3a is attached to the midpoint C of the rod. The other end of the string is attached to a point D on the wall, vertically above A. The rod is in equilibrium when the angle between the rod and the wall is θ, where tan⁡θ=32. A particle of weight W is attached to the point E on the rod, where the distance AE is equal to ka (3<k<6) (see diagram). The rod and the string are in a vertical plane perpendicular to the wall. The coefficient of friction between the rod and the wall is 13. The rod is about to slip down the wall.

Figure 1 of 1 for question 4
a

Find the value of k.

[5]
b

Find, in terms of W, the magnitude of the frictional force between the rod and the wall.

[2]
5

A particle P is projected from a point O on horizontal ground with speed u at an angle θ above the horizontal, where tan⁡θ=13. The particle P moves freely under gravity and passes through the point with coordinates (3a,45a) relative to horizontal and vertical axes through O in the plane of the motion.

a

Use the equation of the trajectory to show that u2=25ag.

[2]
b

At the instant when P is moving horizontally, a particle Q is projected from O with speed V at an angle α above the horizontal. The particles P and Q reach the ground at the same point and at the same time. Express V2 in the form kag, where k is a rational number.

[6]
6

A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held with the string taut and the string makes an angle θ with the downward vertical through O. The particle P is projected at right angles to the string with speed 1310ag and begins to move downwards along a circular path. When the string is vertical, it strikes a small smooth peg at the point A which is vertically below O. The circular path and the point A are in the same vertical plane. After the string strikes the peg, the particle P begins to move in a vertical circle with centre A. When the string makes an angle θ with the upward vertical through A the string becomes slack (see diagram). The distance of A below O is 59a.

Figure 1 of 1 for question 6
a

Find the value of cos⁡θ.

[6]
b

Find the ratio of the tensions in the string immediately before and immediately after it strikes the peg.

[4]
7

A particle P of mass m kg is held at rest at a point O and released so that it moves vertically under gravity against a resistive force of magnitude 0.1mv2 N, where v ms−1 is the velocity of P at time t s.

a

Find an expression for v in terms of t.

[6]
b

The displacement of P from O at time t s is x m. Find an expression for v2 in terms of x.

[5]

Mark scheme

9231/32 · Oct/Nov 2024

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1 · 3 marks

  • B1

    Hooke’s law: T=1000.8x

  • B1

    N2L: T=2×(0.8+x)×52

  • B1

    Equate and solve: 50(0.8+x)=125x,x=0.533

2 · 4 marks

  • B1

    Extension when P comes to rest is x, EPE loss =12×5mga×((a2)2−x2)

  • M1

    Energy: 12×5mga×((a2)2−x2)=mg(a2−x)

  • M1

    Solve: 20x2−8ax−a2=0

  • A1

    x=−a10, so distance of P below O is 910a

3a · 3 marks

  • B1

    PCLM along line of centres: −mv=m2ucos⁡θ−m3usin⁡θ

  • M1

    NEL: v=eu(3sin⁡θ+2cos⁡θ)

  • A1

    Eliminate v: 6sin⁡θ=8cos⁡θ,tan⁡θ=43

3b · 2 marks

  • M1

    Only change in KE is along line of centres Loss = 12m((2ucos⁡θ)2+(3usin⁡θ)2)−12mv2

  • A1

    12mu2(3625+14425−(−125−65)2)=7225mu2

3c · 2 marks

  • M1

    [Components of velocity of A after collision are ←65u↓85u so] angle between line of centres and A’s direction is θ.

  • 1 mark

    Angle of deflection = 180∘−2tan⁡−1(4/3)=73.7∘

4a · 5 marks

  • B1

    ↑Tcos⁡θ+F=2W

  • B1

    →Tsin⁡θ=R

  • M1

    Moments about A: Tcos⁡θ×3asin⁡θ+Tsin⁡θ×3acos⁡θ=W×3asin⁡θ+W×kasin⁡θ OR Moments about C: R×3acos⁡θ=F×3asin⁡θ+W×(ka−3a)sin⁡θ

  • M1

    [6aTcos⁡θ=(3+k)aW and T3(3cos⁡θ+sin⁡θ)=2W give] [and giveand give] 12cos⁡θ=(3+k)(cos⁡θ+13sin⁡θ)

  • A1

    k=5

4b · 2 marks

  • M1

    A complete method to find F in terms of W

  • A1

    F=23W

5a · 2 marks

  • M1

    Use correct equation of trajectory: 45a=3a×13−g2u2×(3a)2×(1+19)

  • A1

    45a=a−5ga2u2,5gau2=15,u2=25ga

5b · 6 marks

  • B1

    For P, time of flight T and range R For Q, time of flight ½ T and range R [T=2usin⁡θg=10ag]

  • B1

    [From motion of P, R=2g×25ag×310=] 15a

  • M1

    For Q: →R=vcos⁡α×12T,vcos⁡α=2RT

  • M1

    ↑0=vsin⁡α×12T−12g(12T)2,vsin⁡α=14gT

  • M1

    Square and add: v2=(2RT)2+(14gT)2[=90ag+58ag]

  • A1

    v2=7258ag

6a · 6 marks

  • 1 mark

    For P to lowest point L: Energy: 12mv2=12mu2+mga(1−cos⁡θ) [v2=289ag−2agcos⁡θ]

  • 1 mark

    From L to string goes slack: Energy: 12mv2=12mw2+4a9mg(1+cos⁡θ) [w2=209ag−269agcos⁡θ]

  • A1

    Both equations correct, allow unsimplified.

  • B1

    When string goes slack: mgcos⁡θ=mw24a9

  • 1 mark

    Equate expressions for w2 to find a value for cos⁡θ

  • A1

    cos⁡θ=23

6b · 4 marks

  • M1

    Tension before: T1−mg=mv2a Tension after: T2−mg=mv249a

  • A1

    [v2=169ag,T1=259mg]T2=5mg

  • M1

    Find the other tension (from a valid equation) and find ratio of tensions

  • A1

    Ratio is 5 : 9

7a · 6 marks

  • B1

    mdvdt=mg−0.1mv2

  • 1 mark

    dvdt=10−0.1v2=110(100−v2) dv100−v2=110dt ln⁡(v+1010−v)=2t+A

  • A1

    ln⁡∣v+1010−v∣=2t+A

  • 1 mark

    Use t=0,v=0,A=0

  • M1

    Remove logs to obtain v in terms of t

  • A1

    v=10(e2t−1)e2t+1 aef

7b · 5 marks

  • 1 mark

    vdvdx=110(100−v2) −12ln⁡(∣100−v2∣)=110x+B

  • A1

    −12ln⁡(∣100−v2∣)=110x+B

  • 1 mark

    Use x=0,v=0,B=−12ln⁡100

  • M1

    Remove logs to obtain v2 in terms of x

  • A1

    v2=100(1−e−x5)

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